common eigenvector of a diagonal element cross-section

Denote by $\\mathrm{M}_{n}(\\mathcal{K})$ the set of all $n\\times n$ matrices over $\\mathcal{K}$ . Let $d_{i}\\colon\\mathrm{M}_{n}(\\mathcal{K})\\longrightarrow\\mathcal{K}$ bethe function which extracts the $i$ th diagonal element of a matrix,and let $\\varepsilon_{i}\\colon\\mathcal{K}^{n}\\longrightarrow\\mathcal{K}$ bethe function which extracts the $i$ th of a vector. Finally denote by $[n]$ the set $\\{1,\\ldots,n\\}$ .

Theorem 1.

Let $\\mathcal{K}$ be a field.For any sequence $A_{1},\\ldots,A_{r}\\in\\mathrm{M}_{n}(\\mathcal{K})$ of uppertriangular pairwise commuting matrices and every row index $i\\in[n]$ , there exists $\\mathbf{u}\\in\\mathcal{K}^{n}\\setminus\\{0\\}$ such that

A_{k}\\mathbf{u}=d_{i}(A_{k})\\mathbf{u}\\quad\\text{for all \\(k\\in[r]\\).}

(1)

Proof.

Let $\\lambda_{k}=d_{i}(A_{k})$ for all $k\\in[r]$ , so that theproblem is to find a common eigenvector $\\mathbf{u}$ of $A_{1},\\ldots,A_{r}$ whose corresponding eigenvalue for $A_{k}$ is $\\lambda_{k}$ . It issufficient to find such a common eigenvector in the case that $i=n$ is the least $i\\in[n]$ for which $d_{i}(A_{k})=\\lambda_{k}$ for all $k\\in[r]$ , because if some smaller $i$ alsohas this property then one can solve the corresponding problem forthe $i\\times i$ submatrices consisting of rows and columns $1$ through $i$ of $A_{1},\\ldots,A_{r}$ , and then pad the common eigenvectorof these submatrices with zeros to get a common eigenvector of theoriginal $A_{1},\\ldots,A_{r}$ .

By the existence of acharacteristic matrix of a diagonal elementcross-section (http://planetmath.org/CharacteristicMatrixOfDiagonalElementCrossSection)there exists a matrix $C$ in the unital algebragenerated by $A_{1},\\ldots,A_{r}$ such that $d_{i}(C)=1$ if $d_{i}(A_{k})=\\lambda_{k}$ for all $k\\in[r]$ , and $d_{i}(C)=0$ otherwise; in other words that matrix $C$ satisfies $d_{n}(C)=1$ and $d_{i}(C)=0$ for all $i<n$ . Since it is also upper triangular itfollows that the matrix $I-C$ hasrank (http://planetmath.org/RankLinearMapping) $n-1$ , so the kernel of thismatrix is one-dimensional. Let $\\mathbf{u}=(u_{1},\\ldots,u_{n})\\in\\ker(I-\obreak C)$ be such that $u_{n}=1$ ; it is easy to see thatthis is always possible (indeed, the only vector in this nullspacewith $n$ th $0$ is the zero vector). This $\\mathbf{u}$ is thewanted eigenvector.

To see that it is an eigenvector of $A_{k}$ , one may first observethat $C$ commutes with this $A_{k}$ , since the unital algebra ofmatrices to which $C$ belongs iscommutative (http://planetmath.org/Commutative). This implies that $A_{k}\\mathbf{u}\\in\\ker(I-\obreak C)$ since $\\mathbf{0}=A_{k}\\mathbf{0}=A_{k}(I-\obreak C)\\mathbf{u}=(I-\obreak C)A_{k}%\\mathbf{u}$ . As $\\ker(I-\obreak C)$ isone-dimensional it follows that $A_{k}\\mathbf{u}=\\lambda\\mathbf{u}$ for some $\\lambda\\in\\mathcal{K}$ . Since $A_{k}$ is upper triangularand $u_{n}=1$ this $\\lambda$ must furthermore satisfy $\\lambda=\\lambda u_{n}=\\varepsilon_{n}(\\lambda\\mathbf{u})=\\varepsilon_{n}(A_{k}%\\mathbf{u})=d_{n}(A_{k})u_{n}=d_{n}(A_{k})$ , which isindeed what the eigenvalue was claimed to be.∎